$$
\begin{align}
\frac{\mathrm{d}x}{\mathrm{d}y}&=\frac{1-xy}{y^2}\tag{1a}\\
\frac{\frac1y\mathrm{d}u-\frac{u\,\mathrm{d}y}{y^2}}{\mathrm{d}y}&=\frac{1-u}{y^2}\tag{1b}\\
\frac{\frac1y\mathrm{d}u}{\mathrm{d}y}&=\frac{1}{y^2}\tag{1c}\\
\mathrm{d}u&=\frac{\mathrm{d}y}y\tag{1d}\\
u&=c+\log(y)\tag{1e}\\
x&=\frac{c+\log(y)}{y}\tag{1f}
\end{align}
$$
Explanation:
$\text{(1b):}$ $x=\frac uy$
$\text{(1c):}$ add $\frac u{y^2}$
$\text{(1d):}$ separate
$\text{(1e):}$ solve
$\text{(1f):}$ $u=xy$